Steve thinks that he has a fair coin with equal probability of landing head or tails. He is ready to change his mind if in a long enough series of flips the coin will land on the same side. Steve decided that "long enough" means the probability of a fair coin landing the same way is less than 1%. What is the shortest series that will allow Steve to declare that the coin is not fair?
A. 7 flips
B. 8 flips
C. 99 flips
D.101 flips
I think its 99 so C but I'm not sure
(1/2)^n < .01
n log(1/2) < -2
n > -2 / log(1/2)
Wdym by that equation?
1/2 is the probability of a given side
for n consecutive flips with the same result , the probability is (1/2)^n
the criteria for the probability of consecutive identical flips is < 1%
To determine the shortest series that will allow Steve to declare that the coin is not fair, we need to calculate the probability of the coin landing on the same side in a given number of flips.
The probability of getting heads in one flip of a fair coin is 0.5, and the probability of getting tails is also 0.5. Since Steve wants the probability of a fair coin landing on the same side to be less than 1%, we need to find the number of flips for which the probability is less than 0.01.
To calculate the probability of getting the same side in a sequence of flips, we use the formula:
Probability = (1/2)^(number of flips)
Let's check each option:
A. 7 flips: Probability = (1/2)^7 = 0.0078 (0.78%)
B. 8 flips: Probability = (1/2)^8 = 0.0039 (0.39%)
C. 99 flips: Probability = (1/2)^99 = 7.94e-30 (0.000000000000000000000000000008%)
D. 101 flips: Probability = (1/2)^101 = 3.94e-31 (0.0000000000000000000000000000004%)
From the calculations, we can see that the probability of the coin landing on the same side is less than 1% only after 99 flips (option C). Therefore, the correct answer is option C.